Explanation:
Percentage Yield
= (3.37g/3.81g) * 100% = 88.45%.
Therefore 88.45% SO2 is the percentage yield.
Answer:
C would be the right answer
Answer:
The water lost is 36% of the total mass of the hydrate
Explanation:
<u>Step 1:</u> Data given
Molar mass of CuSO4*5H2O = 250 g/mol
Molar mass of CuSO4 = 160 g/mol
<u>Step 2:</u> Calculate mass of water lost
Mass of water lost = 250 - 160 = 90 grams
<u>Step 3:</u> Calculate % water
% water = (mass water / total mass of hydrate)*100 %
% water = (90 grams / 250 grams )*100% = 36 %
We can control this by the following equation
The hydrate has 5 moles of H2O
5*18. = 90 grams
(90/250)*100% = 36%
(160/250)*100% = 64 %
The water lost is 36% of the total mass of the hydrate
Answer is: at higher temperatures reaction will go to the right (forward), more products (C₂H₄ and H₂) will be produce, because this is endothermic reaction (ΔH<span> is positive, </span>energy is consumed) and according Le Chatelier's principle <span>heat is included as a reactant. </span> .
